Gravitational potential energy (GPE) (1 Viewer)

[eyeseeyou]

Is it normal to get a negatie answer when doing GPE questions?

 

[eyeseeyou]

In GPE questions when it asks "calculate the energy________" what are they asking?

Also when they ask for "change in GPE" how do you do those types of questions, I never got them

 

[Green Yoda]

In GPE questions when it asks "calculate the energy________" what are they asking?

Also when they ask for "change in GPE" how do you do those types of questions, I never got them

Energy would be E(p) which is -(Gm1m2)/r
Change in E(p) would be Work Done which is -(Gm1m2)/r(f) - (-(Gm1m2)/r(i))

 

[eyeseeyou]

Energy would be E(p) which is -(Gm1m2)/r
Change in E(p) would be Work Done which is -(Gm1m2)/r(f) - (-(Gm1m2)/r(i))

Wait so energy is GPE?

 

[Green Yoda]

Wait so energy is GPE?

Depends on the question..but GPE is an energy..it is the Gravitational Potential Energy

 

[eyeseeyou]

Depends on the question..but GPE is an energy..it is the Gravitational Potential Energy

Oh whoops my bad

 

[eyeseeyou]

Energy would be E(p) which is -(Gm1m2)/r
Change in E(p) would be Work Done which is -(Gm1m2)/r(f) - (-(Gm1m2)/r(i))

In this formula, what does f(f) and r(i) represent? (I know it represents initial and final but initial and final of what)

 

[eyeseeyou]

When it says "calculate the gravitational force that the planet exerts on the satellie" do we just use the GPE formula?

 

[Green Yoda]

In this formula, what does f(f) and r(i) represent? (I know it represents initial and final but initial and final of what)

r(f) and r(i) are the Final and Initial distance of the work done or the change in GPE of the object.

 

[Green Yoda]

When it says "calculate the gravitational force that the planet exerts on the satellie" do we just use the GPE formula?

F(g)=-(Gm1m2)/r^2

 

[eyeseeyou]

BTW rathin have you done projectile motion yet?

 

[Green Yoda]

BTW rathin have you done projectile motion yet?

Started a bit of it

 

[eyeseeyou]

Started a bit of it

What option r u doing?

 

[Green Yoda]

What option r u doing?

Q2Q, lets not spam this thread

 

[eyeseeyou]

How the hell do you do this Q

An object (O) orbiting a planet (P) at an altitude of 100km required 10MJ to move to an altitude of 200km above planet's surface. Planet (P) has an average diameter of 5000 km. If the object then falls back 50km to an altituded of 150km above the surface of the planet, calculate the change in GPE of the fall

 

[Green Yoda]

r=2.5x10^6m + the altitude given

If we rearrange the ΔGPE formula we get E(p)=Gm1m2(1/r(i) - 1/r(f))
∴1x10^7=Gm1m2(1/(2.6x10^6) - 1/(2.7x10^6))
∴Gm1m2=7.02x10^14
Now we calculate the ΔGPE for the fall by subbing in Gm1m2 as 7.02x10^14
∴ΔGPE=7.02x10^14(1/(2.7x10^6) - 1/(2.66x10^6))
∴ΔGPE=-4.9x10^6 J

 

[eyeseeyou]

How do I calculate the period

Also does Period represent gravitational constant in the orbital velocity formula?

 

[eyeseeyou]

r=2.5x10^6m + the altitude given

If we rearrange the ΔGPE formula we get E(p)=Gm1m2(1/r(i) - 1/r(f))
∴1x10^7=Gm1m2(1/(2.6x10^6) - 1/(2.7x10^6))
∴Gm1m2=7.02x10^14
Now we calculate the ΔGPE for the fall by subbing in Gm1m2 as 7.02x10^14
So ΔGPE=7.02x10^14(1/(2.7x10^6) - 1/(2.66x10^6))
ΔGPE=-4.9x10^6 J

What?

 

[Green Yoda]

What?

Diameter is 5000km
Radius is 2500km
Radius is 2500000m
Radius is 2.5x10^5m

 

[Green Yoda]

Is my answer right?